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Question: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of ...

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R2R.
A. 5GmM6R\dfrac{{5GmM}}{{6R}}
B. GmM2R\dfrac{{GmM}}{{2R}}
C. GmM3R\dfrac{{GmM}}{{3R}}
D. 5GmM7R\dfrac{{5GmM}}{{7R}}

Explanation

Solution

Recall that whenever a body is accelerated from zero, then work is needed to be done. The kinetic energy is the energy that arises due to the object in motion. While the potential energy is the energy stored in a body due to its height. Since the satellite is approaching in an upward direction, it will have some gravitational potential energy.

Step-By-Step Explanation:
Step I:
Given that the mass of the satellite is =m = m
And the mass of the planet is =M = M
Radius =R = R
Altitude=2R = 2R
And h is the height
The gravitational potential energy is given by E1=−GmR{E_1} = \dfrac{{ - Gm}}{R}
But the potential energy at altitude is given by −GmMR+2R=−GmM3R - \dfrac{{GmM}}{{R + 2R}} = - \dfrac{{GmM}}{{3R}}
Step II:
The velocity at which a satellite revolves around another body is known as its orbital velocity. The orbital velocity of a satellite is given as
v0=GmMR+h{v_0} = \dfrac{{GmM}}{{R + h}}
The total energy of the system is the sum of its kinetic and potential energy. Therefore,
E2=K.E.+P.E.{E_2} = K.E. + P.E.
E2=12mv02−GmM3R{E_2} = \dfrac{1}{2}mv_0^2 - \dfrac{{GmM}}{{3R}}
Substitute the value of orbital velocity in the above equation and solve,
E2=12GmM3R−GmM3R{E_2} = \dfrac{1}{2}\dfrac{{GmM}}{{3R}} - \dfrac{{GmM}}{{3R}}
E2=GmM3R[12−1]{E_2} = \dfrac{{GmM}}{{3R}}[\dfrac{1}{2} - 1]
E2=GmM3R(−12){E_2} = \dfrac{{GmM}}{{3R}}(\dfrac{{ - 1}}{2})
E2=−GmM6R{E_2} = - \dfrac{{GmM}}{{6R}}
Step III:
The energy required to launch the satellite is given by=E2−E1 = {E_2} - {E_1}
Or −GmM6R−(−GmMR) - \dfrac{{GmM}}{{6R}} - ( - \dfrac{{GmM}}{R})
Or −GmM+6GmM6R\dfrac{{ - GmM + 6GmM}}{{6R}}
Or 5GmM6R\dfrac{{5GmM}}{{6R}}
Step IV:
Therefore the minimum energy required to launch a satellite is 5GmM6R\dfrac{{5GmM}}{{6R}}
⇒\RightarrowOption A is the right answer.

Note: It is to be noted that the gravitational potential energy is sometimes negative. The gravitational potential energy is negative because there exists attractive forces between the bodies. As one of the bodies approaches the other body, the distance between them decreases. The force of attraction exists only in the bodies with opposite charges. Therefore, the gravitational potential energy of the satellite is taken as negative.